Experiment 1
A Review of Mathematical Concepts and Tools
SUMMARY: In Experiment #1, “A Review of Mathematical Concepts and
Tools,” we review mathematical techniques that are commonly used in physics
and astronomy. These include familiar concepts such as exponential notation,
significant figures, and angular measure, as well as concepts such as order
of magnitude calculations and basic statistical measures. Several of the
techniques are combined to analyze the impending collision between the
Milky Way and Andromeda galaxies.
LEVEL OF DIFFICULTY: Moderate
EQUIPMENT NEEDED: Calculator.
L.M. Golden, Laboratory Experiments in Physics for Modern Astronomy:
With Comprehensive Development of the Physical Principles,
DOI 10.1007/978-1-4614-3311-8_1, # Springer Science+Business Media New York 2013
3
4
A.
1 A Review of Mathematical Concepts and Tools
Introduction
From astronomy to business, as well as in many other fields, certain mathematical
tools are of great advantage. They can simplify calculations, prevent errors, and
yield quick estimates. Arithmetic operations such as conversion of units and the
calculation of percentage errors, while difficult for some, are also important to
master for success in many fields.
In this experiment, these and other mathematical tools will be presented. You
will achieve facility in their use by their repeated use in subsequent experiments.
B. Scientific Notation
In many of the sciences, in particular physics and astronomy, we deal with very
small or very large numbers. For example, the largest galaxy in the Local Group
of galaxies, of which the Milky Way is a member, is Andromeda, or M31.
The distance in kilometers to M31 can be calculated by multiplying the number
of light years (ly) to M31, 2.25 million ly, by the number of kilometers in a light
year, 9.46 trillion km/ly,
distance to M31 ¼ 2; 250; 000 ly 9; 460; 000; 000; 000 km=ly
¼ 21; 300; 000; 000; 000; 000; 000 km:
In astronomy, we often use the unit of length known as the parsec. One parsec,
abbreviated as pc, is approximately equal to 3.26 ly. One million parsecs is
abbreviated as Mpc. The values of the light year and parsec, as well as other
physical constants and astronomical measurements, are provided in Appendix I.
We do not want to be encumbered by such calculations. They are time consuming
and we are likely to make errors in carrying the large number of zeroes. In practice,
our calculators will run out of display window space. We run into similar problems
in calculations dealing with very small numbers.
We, accordingly, desire a shorthand symbolism to deal with such very large and
very small numbers. The symbolism which has been adopted by scientists is that of
scientific notation. In scientific notation, numbers are represented by three parts,
a numerical part with a value between 1 and 10, the number 10, and an exponent to
which 10 is raised. A number represented in scientific notation therefore always
takes the form
□
□ 10 .
The exponent locates the decimal point. It tells you the number of places to move
the decimal point to convert the number expressed in scientific notation to ordinary
decimal form. If the number to be represented by scientific notation is greater
than 1, then the exponent is positive because positive exponents tell us to move
B. Scientific Notation
5
the decimal point to the right. If the number to be represented in less than 1, then the
exponent is negative because negative exponents tell us to move the decimal point
to the left.
Thus,
108 ¼ 101 101 101 101 101 101 101 101 ¼ 100; 000; 000
and
108 ¼ 101 101 101 101 101 101 101 101
¼ 0:1 0:1 0:1 0:1 0:1 0:1 0:1 0:1 ¼ 0:00000001:
In the M31 example above, because 12 figures lie to the right of the decimal
point, the number of kilometers in a light year would be represented in scientific
notation as
1 light year = 9:46 1012 km:
The mass of the hydrogen atom, 0.000000000000000000000000001673 kg,
is represented in scientific notation as
mH ¼ 1:673 1027 kg:
This tells us that 27 figures would lie to the left of the decimal point if mH were
expressed as a decimal fraction. Accordingly, the exponent we write is 27.
Now that we have agreed to use this symbolism for expressing small and
large numbers, we can appreciate its usefulness. First, it is much neater and requires
less writing than if we write out the numbers in decimal form. Second, using
scientific notation facilitates arithmetic. When multiplying or dividing numbers
including exponents, we simply add the exponents. It is easy to multiply and
divide numbers between 1 and 10. Third, because it is easy to perform arithmetic
on numbers between 1 and 10, we can avoid errors. Again, it is easy to compare
numbers when expressed in scientific notation. Just look at the exponents. Fifth,
and one of the most important advantages of using scientific notation to
experimentalists, it allows you to clearly express the number of “significant figures”
in a result. Finally, use of scientific notation makes it easy to make “order-ofmagnitude” calculations.
As an example of the advantage of using scientific notation in performing
arithmetic, say we wish to find the result of 230,000,000 190,000/67,000. Converting to scientific notation, this becomes 2.3 108 1.9 105/(6.7 104).
We then combine all the numbers between 1 and 10, and we combine all the
exponents, giving (2.3 1.9/6.7) 108+54. Performing the arithmetic then easily
gives the result, 0.65 109. Because the number preceding the power of 10 is not
between 1 and 10, this is not yet in scientific notation, and we have one more
operation to perform, yielding 6.5 108.
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1 A Review of Mathematical Concepts and Tools
Adding and subtracting numbers expressed in scientific notation can only be
done if the exponent portions are equal. Say, for example, we wish to subtract
3.11 105 from 8.23 107. These must first be expressed with equal exponents.
8:23 107 3:11 105 ¼ 8:23 107 0:0311 107
¼ ð8:23 0:03Þ 107
¼ 8:20 107 :
Any confusion in such arithmetic could always be resolved by simply writing the
number out without scientific notation, although that defeats the purpose of this
convenient shorthand.
As we will see in the following, scientific notation is our friend.
C.
Significant Figures
When you perform a measurement, the precision of your measurement depends on
your equipment. If you measure, for example, the length of a table with equipment
of different precision, you might get 2.0 m, 2.043 m, or 2.0433604 m. The table is
the same. What has changed is the number of digits in which you have confidence,
two, four, and eight, in these cases. Scientists refer to those digits as the number of
significant figures in the measurement. They are the number of digits needed to
express a number to display the precision of its measurement.
(Whenever you write a decimal fraction of value less than 1, always place a
preceding zero to locate clearly the location of the decimal point. Do not write .44;
write 0.44 instead.)
In a measurement, the uncertainty of the final digit can be considered to be +0.5
to 0.5. For example, a measurement of 2.043 m means the real length of the table
is between 2.0425 m and 2.0435 m.
When you are recording data, you should include a final estimated figure beyond
the precision of the measuring instrument, even it happens to be zero. If your ruler
can measure only to 1 mm, for example, estimate the value of the next, uncertain,
digit as well as you can.
Because of ambiguities in the interpretation of the number “zero,” we express
numbers in scientific notation to clearly display their number of significant figures.
Zeroes to the left of non-zero digits are not significant. In 0.000386, only the 3, 8,
and 6 are significant. To express this clearly, we can rewrite this number as
3.86 104. The number is easily seen to have three significant figures. No
ambiguity is present in this case.
Zeroes to the right of non-zero digits, however, present a problem. In the number
9340400, we do not know if the final zeroes are significant. They are needed to
place the decimal point, but they may also be significant. They are only significant
if they are the result of the measurement.
C. Significant Figures
7
If we rewrite the number as 9.34 106, however, then we are stating that we
have three significant figures. If we rewrite it as 9.340 106, then we are stating
that we have four, and if we rewrite it as 9.340400 106 then we are stating that we
have seven significant figures. In this case, then, use of scientific notation unambiguously communicates the number of significant figures.
When we combine one or more measured quantities in a calculation, we also refer
to the number of significant figures in the calculated result. Specific common-sense
rules guide us in determining the number of significant figures in the result. In adding
and subtracting numbers, drop all the digits beyond the first uncertain figure. For
example, let us add 14.49, 7.99833, and 0.2631. Since only 14.49 is known to
hundredths, it makes no sense to add the digits beyond that place. Round the numbers
to the hundredths place, and then save time by dropping all digits beyond that
place before performing the calculation. Our result for the sum is 22.75. In general,
then, do not carry the result beyond the first digit containing an uncertain figure.
In multiplying and dividing, the result should have the same number of significant figures as the term with the fewest. If we multiply 1.78 14.339 and ignore
the significant figures, we will calculate 25.52342. Only the first three digits
are significant, however, so that the answer should be expressed as 25.5. To quote
more significant figures gives a false impression of the precision of the
measurements and your confidence in the final calculated result.
In such calculations, do not confuse the number of significant figures of
constants with those of measured quantities. The former have no bearing on the
number of significant figures in the calculated result. For example, the number p
is known to be 3.14159265. . .. The number of significant figures in a calculation
involving p is only determined by the precision in the measured quantities.
In the calculation of the circumference of a circle, C, from its radius, r, C ¼ 2pr.
The presence of p does not mean that the calculated result has nine or more
significant figures. The presence of the 2 does not mean that the calculated result
has only one significant figure.
As a final comment, “precision” should be distinguished from “accuracy.”
Precision refers to the number of significant figures in a number. Accuracy refers
to the agreement between a number and the actual magnitude of the entity being
measured. Inaccurate results often result from the presence of systematic as
opposed to random errors.
The two should not be confused. For example, if we have a table which is known
to be 3.11 m long, then a measurement of 3 m would be an accurate measurement of
its length with low precision. A measurement of 3.1 m would be an accurate
measurement of its length with greater precision. A measurement of 4.015832 m,
on the other hand, would be an inaccurate measurement of its length quoted with
great precision.
A famous anecdote illustrates this difference. The people of an ancient Chinese
dynasty, who were forbidden to gaze upon the emperor, were asked to guess his
height. After thousands were polled, the height of the emperor, obtained by
8
1 A Review of Mathematical Concepts and Tools
averaging all the responses, was announced to be (let’s say) 5.840273 ft. Of course,
despite the precision of this result it lacked accuracy, none of the people having ever
seen the emperor.
Unfortunately, one often finds figures which are quoted to high precision but
which have low accuracy. This is a favorite tactic in politics and advertising.
For example, “78.7% of doctors recommend Sugar Chewie Choco-Bombs to their
patients who chew gum” is more persuasive than “more than 3/4 of all doctors
recommend Sugar Chewie Choco-Bombs to their patients who chew gum.”
D.
Order of Magnitude Calculations
An order of magnitude calculation is a calculation which leads to a result accurate
to one significant figure. It is performed using scientific notation, and the exponent
to which the 10 is raised is referred to as the “order of magnitude” of the result. As a
result, these calculations could also be considered “factor of 10” calculations. When
faced with a completely unfathomable problem, instead of making a wild guess or
relying on authority, faith, revelation, or bombast to impose an answer, this
technique can produce a meaningful estimate.
Making order of magnitude calculations is valuable not only in science but also
in many other disciplines, often being the only calculation that can be made. We can
transform a state of complete ignorance to a state of reasonable knowledge. An order
of magnitude calculation can settle disputes, aid in designing an experiment, help in
estimating costs, or allow evaluation of a suggested hypothesis.
In this technique, we replace the difficult problem of estimating the value of
some highly unknown quantity with the more manageable problem of estimating a
number of others, for each of which a reasonably accurate estimate can be determined by everyday experience, common sense, or quick reference. We multiply
these estimated factors together and more or less hope that the various errors will
balance each other, leading to a result for the original quantity in which we have
confidence to one significant figure.
For example, let us say we want to estimate the value of a quantity which we
can segment into five factors for each of which we have a reasonably accurate
estimate. We don’t really know the errors in the various factors, that implying
that we in fact know their true values. Then, if the first factor is incorrect by
being a factor of 2 too small, the second factor is incorrect by being a factor of
10 too large, the third is incorrect by being a factor of 4 too small, the fourth factor
is incorrect by being a factor of 2 too large, and the fifth is incorrect by being
a factor of 5 too small, when multiplied together the final result will be incorrect
by a factor of 2 1/10 4 1/2 5 ¼ 2, a remarkable achievement. The more
factors involved the better the chance that the errors will cancel and that the
estimated value will be close to the actual value. The technique will work if about
D.
Order of Magnitude Calculations
9
as many of the individual estimates are incorrect by being too large as are
incorrect by being too large.
Before you begin, however, you should have a reasonable idea as to the kinds
of values you might expect. If you are estimating the number of people in California
who weigh more than 300 lb., you know that an answer of 5 or 100,000,000 will be
wrong. If you are estimating the number of $1 bills in circulation, you know that an
answer of 7 or 4 1011 can’t be correct. Making this initial intelligent guess helps
ensure that your result makes sense.
Let us, for an illustration, try, to estimate the number of grains of sand on the
coastlines of the Earth. “Impossible!,” you say. Don’t be so sure. To do this we
need to know the number of grains of sand in a cubic volume, say a cubic
centimeter, and the total volume of coastline sand on Earth. Pick up a handful of
sand. One inch equals 2.54 cm, so a centimeter is about the size of a fingernail.
Let us say you can place 30 grains of sand along your fingernail. Then the number
of grains of sand in 1 cm3 is 30 30 30 ¼ 2.7 104 grains/cm3. Because the rest
of our calculations will be done using kilometers, let us convert this result using
1 km3 ¼ 1015 cm3. That is, 2.7 104 g/cm3 ¼ 2.7 1019 grains/km3. We might
believe that this number is accurate to within a factor of 10.
Now, to find the volume of sand in the world, we can start with the circumference of the Earth, about 40,000 km (25,000 miles). Although we might be able to
find this information in an encyclopedia, let’s say that the length of coastline is
about 20 times the circumference of the Earth, 20 40,000 km ¼ 8.0 105 km.
For the width of sand along a typical coastline take 10 m, and for the depth take 1 m.
Although these are simply estimates from our own experience, we believe that they
are accurate to factors of 10. The typical width of a coastline covered with sand is
not, that is, closer to 100 m or 1 m than it is to 10 m, and the typical depth of sand
is not closer to 0.1 m (about 4 in.) or 10 m (about 33 ft) than it is to 1 m.
To obtain our order of magnitude estimate of the number of grains of sand on the
coastlines of the world, we then multiply these various factors.
# of grains of sand ¼ ð# of grains in 1km3 Þ ðvolume of sand in km3 Þ
¼ ð2:7 1019 grains=km3 Þ ½ð8:0 105 kmÞ ð102 kmÞ ð103 kmÞ
¼ 2 1020 grains of sand:
Note that the final result is rounded to one significant figure. Because of the
hopeful balancing of the various errors in our estimates, we believe that this result
is accurate to within an order of magnitude. The true number of grains on the
coastlines of Earth, therefore, we believe to be roughly between 2 1019 and
2 1021. (An estimate of the number of stars in the universe, 300 billion stars per
galaxy times a billion galaxies or 3 1020 stars, comes out to about this same
number, a useless if highly inconsequential fact.)
As is clear, performing order of magnitudes calculations is somewhat of an art.
No such thing as a correct answer exists.
10
1 A Review of Mathematical Concepts and Tools
E. Conversion of Units
Conversion of units, although the source of much anguish, can be done easily using
a simple rule: We can multiply or divide any number by 1. To convert units, take the
conversion formula, divide one side by the other, and then multiply or divide the
number to be converted by this quotient.
For example, let us say we want to convert the diameter of the Earth in
kilometers, 12,756 km, to miles. We know,
1 mile = 1.61 kilometer:
Dividing one side of this equation by the other,
1 mile/1.61 km = 1:
To convert 12,756 km to miles we can multiply or divide by 1. The choice is
determined by our desire to cancel out the unwanted unit, in this case kilometer
(written in bold).
12,756 km 1 ¼ 12; 756 km (1 mile/1:61 kmÞ
¼ 7923 miles:
This recipe can be used in the conversion of any units.
As with order of magnitude calculations, you should have a reasonable idea of
the final result before you begin the conversion. That will be a guide as to whether
your result is sensible. In this way, you know that 400 miles cannot be the
equivalent of 3 km or 75,000 km. A decent guess might be between 100 and
1000 km.
F. Calculation of Errors
1. Percentage Errors
In experiments we often want to find the percentage error between a measurement
and a known value or a percentage difference between two quantities. If x is the
measured value of a quantity which is known to have a value of s, then the
percentage error is
xs
percentage error ¼ 100:
s
G.
Mean and Standard Deviation
11
If we are comparing quantity x1 to quantity x2, then the percentage difference
between them is
x1 x2
percentage difference ¼ x2
100:
In some situations, we can estimate the error in a measured quantity and wish to
then calculate the corresponding estimated percentage error in the measured quantity. If Dx is the estimated error in a measured quantity xo, then,
Dx
estimated percentage error ¼ 100:
xo
The Greek capital letter delta, D, is used to denote differences in quantities.
In all these calculations, because percentages can only be positive, we calculate the
absolute values of the differences.
2. Propagation of Errors
Often we encounter a quantity which is the product of more than one variable, each
raised to a different power. If we know the uncertainties in the individual variables,
then we can calculate the uncertainty in the product. In general, if f(x,y) ¼ a xn ym ,
then by taking the differentials and dividing the result by f(x,y) one finds
Df
Dx
Dy
¼n
þm :
f
x
y
This is valid for any values of n and m, including non-integers.
G.
Mean and Standard Deviation
You are most likely familiar with the techniques of calculating the mean and
standard deviation of a group of data. The mean is defined as the sum of the
individual values or measurements, xi, divided by the number of values. If we
have, for example, five measurements, x1, x2, x3, x4, and x5, then the mean is
xav ¼
x1 þ x2 þ x3 þ x4 þ x5
:
5
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1 A Review of Mathematical Concepts and Tools
This can be generalized, using the Greek letter sigma, S, to indicate a
summation,
xav ¼
n
1X
xi ;
n i¼1
where n is the number of individual values and the Greek letter S signifies the sum
of all the values. This is often simply called the “average.”
The standard deviation is a measure of the distribution of those individual values
about the mean. Again using the shorthand summation symbol, it is defined as
2P
n
s ¼ 4i¼1
ðxi xav Þ2
31=2
5
:
n1
For example, let us say we have a set of measurements taken by different people
of the size of a meteorite that we found in the desert. Those measurements are 17.8,
17.2, 18.1, and 17.7 mm. The mean is found to be
xav ¼ ð17:8 þ 17:2 þ 18:1 þ 17:7Þ=4 mm
¼ 17:7 mm;
The standard deviation is calculated to be
n
o1=2
s ¼ ½ð17:8 17:7Þ2 þ ð17:2 17:7Þ2 þ ð18:1 17:7Þ2 þ ð17:7 17:7Þ2 =3
¼ f½0:01 þ 0:25 þ 0:16 þ 0:0=3g1=2
¼ 0.37 mm:
Sometimes the variance is quoted. This is simply the square of the standard
deviation,
n
P
s2 ¼
ðxi xav Þ2
i¼1
n1
:
We may want to calculate the mean of quantities which have different
uncertainties. In that case, we want to give less weight to the less certain quantities
and more weight to the more certain quantities. This is achieved by calculating a
weighted mean. If the weight assigned to measurement xi is wi, then the weighted
mean of the n quantities is
n
P
w i xi
i¼1
xav ¼ P
:
n
wi
i¼1
H.
Angular Measurement
13
The standard deviation of n quantities with weights wi whose mean is xav is
2P
n
6i¼1
s¼6
4
½wi ðxi xav Þ2
n
P
wi
31=2
7
7
5
:
i¼1
H.
Angular Measurement
The hopelessly non-decimal system of angular measure comes to us from Babylonian tradition through centuries of use. A full circle is divided into 360 degrees of
arc, a degree is subdivided into 60 minutes of arc, and a minute of arc is further
subdivided into 60 seconds of arc. For degrees, minutes, and seconds we use the
symbols , 0 , and 00 . (The Babylonians used the sexagesimal system, the base of
their counting being 60 rather than our 10. They also knew that the perimeter
of a hexagon is exactly equal to six times the radius of the circumscribed circle.
The number 6 60 ¼ 360 is thereby associated with a circle, and would be a fairly
obvious choice by which to divide the circle if you were a Babylonian.)
An angular size can be given either in these units or in decimal form. For example,
2 300 could be rewritten 2.50 , and 270 2500 could be rewritten 27.420 .
We sometimes need to convert between degrees, minutes of arc, and seconds
of arc and degrees and decimal fractions of a degree. For example, to express 27.14
in degrees, minutes of arc, and seconds of arc we note that 0.14 is the same as
0.14 60 (minutes of arc/degree) ¼ 8.40 . Then we note that 0.40 is the same
as 0.40 60 (seconds of arc/minute of arc) ¼ 2400 .
To transform from degrees, minutes of arc, and seconds of arc to degrees and
decimal fractions of a degree, we perform an addition. For example,
0
00
15
18
þ
0
00
60 per deg ree ð60 60Þ per deg ree
15
18
þ
¼ 64 þ
60
3600
64 15 18 ¼ 64 þ
¼ 64 þ 0:25 þ0:005
¼ 64:255 :
Because the system of degrees, minutes, and seconds is essentially arbitrary,
it should be no surprise that it cannot be employed in the trigonometric calculations developed independently by the Greeks. They discovered that the ratio
of the circumference, C, of any circle to its diameter, D, is the number
14
1 A Review of Mathematical Concepts and Tools
p ¼ 3.14159. . . 1 That is, C ¼ pD. We frequently rewrite this in terms of the radius R
of the circle, C ¼ 2p R.
To determine the kind of angular measure that must be employed in trigonometric calculations, examine a circle. In particular, what is the portion, s1, of the
circumference of a circle subtended by an angle of 1 ? By a simple proportion,
1
s1
;
¼
360
2pR
or
s1 ¼
2pR
:
360
In fact, this result is entirely general for any angle, y , in degrees, subtending any
portion of circumference, s,
y
s
¼
;
360 2pR
or
s¼
2pR
y:
360
(1)
For y ¼ 360 , s ¼ C ¼ 2pR, as it must.
Now, note that we can rewrite (1) as
s¼
y
R:
ð360=2pÞ
This tells us that if we express y in units, not of degrees, but in units of some
funny number 360/(2p), then we can write the portion of circumference simply,
without regard to any arbitrary Babylonian construct,
s ¼ Ry;
where now y is in units of
(2)
360
2p .
1
The century-old anecdotal story that Johann Strauss, Jr., (1825–1899) composed the famous Blue
Danube Waltz while eating “pies” and therefore decided to denote that work as his “opus 314”
apparently was a hoax perpetrated by classical music-loving geometry theorists with an addiction
to apples. On the other hand, perhaps some music-loving astronomer made the whole thing up.
I.
Scale Factors
15
Fig. 1 The size of an object
can be determined if its
distance and angular size
are known
Because of the importance of this number and its intimate association with the
radius of the circle, it is given a name, radian. One radian is a unit of angular
measurement equal to 360
2p ¼ 57.3 . Angles given in radians are said to be expressed
in circular measure. The lack of any constants in (2) tells us that this is the natural
unit for angular measure and, accordingly, the natural unit for trigonometry.
Note that the strange value for the unit of radian is not its fault. Nature made the
trigonometry of circles so that C ¼ 2pR. The arbitrary (except to the Babylonians)
division of a circle into 360 parts determines the value of 57.3 .
Equation 2 is related to an important approximation that we will encounter
frequently, the small angle approximation. In (2), s is a portion of an arc length.
If R s, that is, for objects at comparatively great distances, the curvature of the
circular arc can be neglected and s can be considered a linear length. In general, given
an object of measured angular size and known distance, as shown in Fig. 1, we
calculate the size of the object from tan y/2 ¼ s/(2R). If R s, however, we can use
the small angle approximation for tangent to find s ¼ R y, which is (2). This condition
is, of course, frequently the case in astronomical observations. To apply (2), the
angular size of the object must be measured in radians.
I.
Scale Factors
Scale factors are one of those concepts that are familiar to everyone, but when placed
before students can cause consternation. We are all familiar with the scale of a map.
The distance from Chicago to Springfield, Illinois, is about 180 miles. On a road map,
with a scale of 20 miles to 1 in., the distance on the map is about 9 in.
In astronomy, we often have to determine the scale of spectra or photographs of
star fields or galaxies. As a road map spans a range of miles, so a spectrum spans a
range of wavelengths and a photograph spans a range of seconds or minutes of arc.
To determine the scale of a road map, we could lay a ruler along the path from
one location to a second whose distance from the first is known. We could then read
off the number of inches between the two locations, divide by the known distance,
and then calculate that the scale of the map is so many miles per inch,
fmap ¼
DD
;
DL
16
1 A Review of Mathematical Concepts and Tools
where DD is the distance between the locations in miles and DL is the number of
inches between them on the map. Henceforth, when we want to find the number
of miles between two locations, we measure the number of inches and multiply by
this scale factor.
With a spectrum or photograph, the procedure is exactly the same. To determine
the scale factor of a spectrum, we lay a ruler between two spectral lines each of
whose wavelengths is known, measure the number of millimeters between them,
and then find the scale factor by calculating
fspectrum ¼
Dl
;
DL
where Dl is the wavelength interval between the spectral lines in angstroms, the
unit of wavelength (1 Å ¼ 108 cm), and DL is the distance between them in
millimeters. Henceforth, when we want to find the number of angstroms between
two spectral lines in this spectrum, we measure their separation in millimeters and
multiply by this scale factor.
To determine the scale factor of a photograph, we lay a ruler between two stars
or parts of a galaxy, the angular distance between which is known in seconds or
minutes of arc, measure the number of millimeters between them, and then find the
scale factor by calculating
fphoto ¼
Dy
;
DL
where Dy is the number of seconds or minutes or arc between the two stars or parts
of the galaxy, and DL is the distance between them in millimeters. Henceforth,
when we want to find the angular distance between two locations in this photograph, we measure their separation in millimeters and multiply by this scale factor.
In determining a scale factor, use two points that are as widely separated as
possible. In this way, the errors in reading the ruler will be small compared with the
length being measured.
J. Julian Dates
Astronomers frequently need to determine the time interval between celestial
events, the time interval between the dates of their observations of celestial events,
or to coordinate observations of the same phenomenon, be it solar flares or
supernova explosions, for various examples. Using a calendar poses numerous
problems, such as different number of days in different months and leap years.
In calendars such as ours, division into time periods of different lengths, such as
months and years, causes unnecessary complications. More than that, different
cultures use different calendars and historical events in different calendars can be
J. Julian Dates
17
Fig. 2 The flux density (brightness in radio wavelengths) of thesource 3C273B as a function of
Julian date. Using Julian dates facilitates the determination of time intervals. Observations at the
Hat Creek Radio Observatory (HCRO) by the author
difficult to correlate chronologically. A simpler manner of keeping track of time,
in terms of the number of days in a sequence, was therefore needed. It is simply
more convenient to reckon time in one single unit, be it days or seconds, rather than
days, months and years.
The method of choice among astronomers is the Julian date. It is defined as the
number of days reckoned from 12:00 noon universal time on January 1, 4713
B.C.E. Universal time, or UT. Universal time is the time at the Prime Meridian,
the meridian or line of longitude where the longitude is defined to be 0 . Because
the meridian was chosen to pass through the Royal Observatory at Greenwich,
England, universal time was formerly referred to as Greenwich mean time, or GMT.
The date of January 1, 4713, was chosen as the zero date to commemorate the date
that aliens brought the first recipe for pistachio ice cream to the Earth.
In Julian days, the time during the day is expressed as decimal fractions of a day.
Midnight on January 1, 2000, for example, has a Julian date of 2451544.5. The
Julian date is frequently quoted without the first two digits. J.D. 2500000.0 will
occur on August 31, 2132 at noon UT.
Use of the Julian date greatly simplifies reckoning of time, being a simple
sequence of numbers increasing by unity from day to day. Some labor is required
to calculate the number of days that have passed between, for example, December
4, 2010, and June 18, 2013. Knowing that the respective Julian dates are 55534 and
56461 makes the task one of simple subtraction.
Figure 2 shows actual observations of the quasar 3C273B. This, as other galaxies
with active galactic nuclei, or AGN’s, are notable for many reasons, including the
brightnesses with time. From this graph, we can easily see that the brightness at
22 GHz frequency varies about 25% over a period of less than about 375 days. If the
18
1 A Review of Mathematical Concepts and Tools
data were plotted instead against calendar date, determining the length of time of
this variation would be unnecessarily time-consuming.
For the purposes of the experiments in this book, we won’t bother with
converting local time to strict Julian dates, referred to the Prime Meridian. Instead,
we’ll simply use the Julian date as that at noontime at your particular location, with
its own time zone. This, if the change of date is altered to occur at midnight rather
than at noon, is sometimes referred to as the chronological Julian date.
K.
The Method of Least Squares
In astronomy and physics we often find the need to find the best-fit curve to a set of
data. In general, if y is a function of independent variables x1, x2, x3, . . . xn,
y ¼ ao þ a1 x1 þ a2 x2 þ a3 x3 þ ::: þ an xn ;
(3)
then we wish to determine the value of the various coefficients ao, a1, a2, a3 . . . an.
The preferred method of doing this is the least squares method, based on the
criterion that the square of the deviations of the observed values of y to the curve
determined by the parameters ai is a minimum. This, in fact, is derived from the
maximum likelihood method of statistical analysis.
If (3) is multiplied in turn by 1 and the various values xi , and each of the n + 1
resulting equation is summed over all the observations, we obtain a set of n + 1
equations in n + 1 unknowns. These can then be solved for the values of ai by any of
the well-known methods for solving simultaneous equations.
For the case of a linear equation of one independent variable,
y ¼ a þ bx;
the set of two equations in two unknowns is
n
X
yi ¼ n a þ b
i¼1
n
X
i¼1
xi yi ¼ a
n
X
xi
(4)
i¼1
n
X
i¼1
xi þ b
n
X
xi 2 :
(5)
i¼1
In actual practice, the variables xi may be powers of independent variables,
trigonometric functions of independent variables, or other functions of the independent variables. The least squares fit of Fig. 5 of Experiment #2, “A Review of
Graphing Techniques,” is one example of data fit by this method.
L.
Galaxies Collide!: The Impact of the Milky Way and Andromeda Galaxies
19
L. Galaxies Collide!: The Impact of the Milky Way
and Andromeda Galaxies
To provide an example of how astronomers use mathematical physics to learn
about the universe, and to provide a real-world application of scientific notation,
significant figures, and the conversion of units, we will calculate the velocity at
which M31 will collide with the Milky Way galaxy. This calculation will be based
on the law of conservation of energy and some simplifying assumptions, and will
enable us to determine if an “air bag” will actually become deployed! Currently,
M31 is about 780,000 pc away and moving toward us with a relative velocity of
about 120 km/s.
As it “falls” toward us, M31 gives up some of its gravitational potential energy,
which is transformed into kinetic energy. We learn in physics that gravitational
potential energy is the energy a mass has by virtue of its presence in a gravitational
field and that kinetic energy is the energy a mass has by virtue of its motion.
The law of conservation of energy tells us that the energy of an isolated system
remains constant in time. Simplifying the problem by accounting only for gravitational potential energy and kinetic energy, we therefore equate the energy of the
M31-Milky Way “system” at the present time with that at the time that the collision
occurs,
KEnow þ PEnow ¼ KEimpact þ PEimpact ;
(6)
using the conventional notation KE for kinetic energy and PE for gravitational
potential energy. The well-known formulas for kinetic energy of an object and the
gravitational potential energy between two objects are
1
KE ¼ mv2 ;
2
where m is the mass of the object moving at a velocity v,
and
PE ¼ Gm1 m2
;
r
20
1 A Review of Mathematical Concepts and Tools
where G is the constant of gravitation, G ¼ 6.67 1011 Nt-m2/ kg2, m1 and m2 are
masses of the two objects which are pulling at each other, and r is the distance
between them. (This traditional but at first sight strange formula puts the arbitrary
zero reference point at infinity. Because only changes in energy are relevant, the
reference point can be placed anywhere. Objects whose separation is less than
infinite have smaller, therefore negative, values of gravitational potential energy.)
This equation results directly from Newton’s Universal Law of Gravitation describing the gravitational force between two objects of masses m1 and m2 which are
separated by a distance r,
F¼
Gm1 m2
:
r2
Identify m1 and m2 as the masses of M31 and the Milky Way, respectively.
Then (6) becomes
1
m1 m2 1
m1 m2
m1 v2now G
¼ m1 v2impact G
:
2
2
rnow
rimpact
(7)
We see that the mass of M31, m1, cancels out, and we can solve for the velocity
of impact,
vimpact
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 2
1
1
vnow þ Gm2
:
¼ 1:414
2
rimpact rnow
(8)
For the value of the distance at which impact occurs, rimpact , use the radius of the
Milky Way galaxy, about 15 kiloparsecs.
This irrelevance of the mass of M31 in (7) and (8) is no surprise. Indeed, it is
explained by none other than Sir Isaac Newton and Albert Einstein. Starting
when Newton in 1666 was aroused from whatever were his daydreams by the
falling British apple in that family garden in Woolsthorpe, Lincolnshire, physicists
eventually realized that all objects under a given force of gravity fall with the
same acceleration, independent of their mass. The equality of the inertial mass,
which describes the acceleration of any object under the action of any force via
Newton’s Second Law, F ¼ ma, with the gravitational mass, which describes the
strength, specifically, of the force of gravity acting on an object, was thereby
established. Einstein stated this in 1907 as one version of his equivalence principle.
As a result, on the moon, in a vacuum, and in any other environment lacking the
frictional drag resulting from an atmosphere, a feather, block of lead, or member of
Congress dropped from the same height at the same time will land at the same time.
One of the Apollo 15 astronauts, if any proof was needed, demonstrated this using a
falcon feather and a geological hammer.
It can be shown that the mass in the expression for kinetic energy is the same as
the inertial mass. It follows, therefore, that the masses m1 in the equation appearing
L.
Galaxies Collide!: The Impact of the Milky Way and Andromeda Galaxies
21
either in the expressions for KE or those for PE are equal and will cancel out, as we
found out.
We cannot perform this calculation without first converting some units. We note
that the units of G are in meters and kilograms and the units of velocity are in km/s,
both consistent in the meter-kilogram-second system of units. Unfortunately, here
the mass is expressed in solar masses, the distances are expressed in parsecs and
kiloparsecs, and vnow is expressed in km/s.
We take a value for the Milky Way mass of 580 billion solar masses. Then, using
the conversion factors
1 pc ¼ 3:086 1016 m;
1 solar mass ¼ 1:99 1030 kg;
1 km ¼ 1000 m;
and
1 kiloparsec ¼ 1000 parsecs;
we can convert the Milky Way mass, m2 , to kilograms, rnow and rimpact to meters,
and vnow to meters per second. For purposes of calculation, note that the unit of
Newtons in the meter-kilogram-second (MKS) system of units, abbreviated Nt,
has the following equivalent: 1 Nt ¼ 1 kg-m/sec2 . This unit appears in the value of
the gravitational constant, G. The equivalence follows from its definition via
Newton’s Second Law, F ¼ ma. Accordingly, the units of G can also be given
as m3/kg-sec2.
Then, with all the factors expressed in the meter-kilogram-second system
of units, we will be able to solve the above equation for vimpact in meters per
second.
Thus,
m2 ¼ 5:80 1011 solar masses
¼ 5:80 1011 solar masses 1:99 1030 kg
1 solar mass
¼ 1:15 1042 kg;
rnow ¼ 780; 000 pc
¼ 7:80 105 pc ¼ 2:41 1022 m;
3:086 1016 m
1 pc
22
1 A Review of Mathematical Concepts and Tools
rimpact ¼ 15; 000 pc
¼ 1:50 104 pc 3:086 1016 m
1 pc
¼ 4:63 1020 m;
and
vnow ¼ 120
km
s
km 103 m
s
1 km
m
5
¼ 1:20 10 :
s
¼ 1:20 102
We see the parameters have three significant figures. The results of the conversion of units, therefore, also have three significant figures.
The calculation of (8) is then easily performed using scientific notation,
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
m2
m3
1:20 105
vimpact ¼ 1:414
þ 6:67 1011
1:15 1042 kg:
2
s
kg s2
:
1
1
4:63 1020 m 2:41 1022 m
Note that both terms under the radical sign have the dimensions of m2/s2. In
the second term, we can cancel out the units of kilograms and one of the three units
of meters and perform the arithmetic,
vimpact
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
m2
m2
¼ 1:414 0:720 1010 2 þ 7:67 1031 2 2:12 1021
s
s
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
m2
¼ 1:414 17:0 1010 2
s
m
¼ 5:83 105 :
s
Converting the result to km/s yields vimpact ¼ 583 km/s.
Whether or not a shock wave will be created when the two galaxies collide
depends on the velocity of impact compared to the velocity with which pressure
waves move in the interstellar space of the Milky Way. If the collision velocity is
greater than the speed of the pressure waves, then the pressure waves cannot move
the interstellar material out of the way fast enough to avoid a build-up. That leads to
a shock wave. Sound waves are, in fact, pressure waves, so that another way of
expressing this condition is whether the velocity of impact is greater than the speed
L.
Galaxies Collide!: The Impact of the Milky Way and Andromeda Galaxies
23
of sound in interstellar space. The related interaction between an aircraft and the
atmosphere is described as supersonic, faster than sound.
The velocity of pressure waves in a given medium depends on the temperature of
the medium and the gas under consideration. For hydrogen, the major constituent
of interstellar gas, at a temperature of 10 K the speed of sound is about 150 km/s.
This compares to the speed of sound in air of about 340 m/s, or about 0.34 km/s.
(This explains the time delay between seeing a lightning bolt and hearing the
thunder, the lightning bolt traveling at the speed of light, which is much larger
than the speed of sound in air. By counting the seconds before you hear the thunder,
you can thereby determine an approximate distance to a lightning bolt.)
The impact velocity of 583 km/s is significantly greater than the velocity of
pressure waves, the “speed of sound,” in interstellar space. A significant shock
wave will be created. That shock wave, resulting in a build-up of interstellar gas and
dust, leads to a flurry of star formation. That build-up of matter and the large
amount of radiation emanating from the large number of newly-born stars create
Brewster Rockit’s “air bag”!
Note that if the mass of the Milky Way is in fact smaller than 580 billion solar
masses, the impact velocity will be smaller, whereas if the mass of the Milky Way is
larger than 580 billion solar masses, the impact velocity will be larger. This has a
direct effect on the amount of time it will take for the two objects to collide. (That
research in 2009 showed the mass of the Milky Way galaxy to be indeed greater
than this figure, leading to a decreased time scale for its collision with M31, was
provided to the cartoonist of “Brewster Rockit: Space Guy!” by the author. We also
advised the cartoonist that the Milky Way is now known to be a barred spiral.
Referring back to the cartoon strip, we see that he is obviously a good student!)
24
M.
1 A Review of Mathematical Concepts and Tools
Mathematical Concepts Experiment Exercises
STUDENT’S NAME ________________________________________________
These various mathematical tools will be of great value to you in any quantitative
field, in business, the social sciences, the arts, as well as in the physical sciences.
The following will help you master them. Show all your calculations.
Circle those numbers which are given in proper scientific notation.
1. .11 104
2. 0.11 104
3. 1.1 103
4. 11.0 102
5. 8.9 1017
6. 8.9 1017
7. 8.90416 1017
8. 8.90416 1017
9. 0.6 101
10. 6.0 101
Express the following in scientific notation. Assume that all have four significant
figures.
11. 1,989,000,000,000,000,000,000,000,000,000 kg
12. 299,800,000 m/s
13. 0.00000000006668 Nt-m2/kg2
14. $30,000,000,000
M.
Mathematical Concepts Experiment Exercises
25
STUDENT’S NAME ________________________________________________
Circle the larger of the following pairs of numbers.
15. a) 9.9 102
b) 1.01 107
16. a) 6.6 106
b) 6.6 108
17. a) 1.44897 107
b) 8.4 103
18. a) 5 104
b) 5 104
Perform the following calculations, showing your intermediate steps without using
a calculator. Express the results in scientific notation.
19. (8.2 1068) (2.00 107) ¼
20. (3.0 1016) (6.0 104) ¼
21. (2.2 1052) (5.0 1014) (2.0 1021) ¼
In the following two exercises, each figure is given to two significant figures.
Express the result with the correct number of significant figures, showing your
intermediate steps.
22. 4.2 104 5.2 102 ¼
23. 7.7 1012 + 2.3 1011 ¼
26
1 A Review of Mathematical Concepts and Tools
STUDENT’S NAME ________________________________________________
Express the results of the following calculations with the correct number of
significant figures.
24. 14.448 + 1.89 + 66.0302 ¼
25. 4.4 + 14.332 + 109 ¼
26. 14.339 + 3.14 22.1 ¼
27. 1.119 4.39 ¼
28. 194 22.02 ¼
29. (72.29 + 1.8) 3.039 ¼
M.
Mathematical Concepts Experiment Exercises
27
STUDENT’S NAME ________________________________________________
30. Calculate to the correct number of significant figures from the following
formula the volume of a sphere whose radius r is measured to be 3.08 cm.
4
V ¼ pr 3 ; where p ¼ 3:14159265:
3
31. The velocity of recession v of the most distant objects in the universe, quasars,
v
can be calculated from their Doppler shift by the formula z ¼ Dl
l ¼ c , where Dl
is the Doppler shift of light of wavelength l observed from the quasar, and the
speed of light c ¼ 3.00 105 km/s. (This formula is derived as Eq. (6) of
Experiment #9, “Determination of the Rotation Rate of Planets and Asteroids
by Radar: Part I: Observations of Mercury,”.) If a given quasar has a Doppler
shift of Dl ¼ 583 Å for light of wavelength 3646 Å, calculate its velocity of
recession. Give your answer to the correct number of significant figures.
(1 Å ¼ 108 cm)
32. “Chicago Slim” Golden (who famously stated “da only famous card-counters
are da ex-card-counters”) spends 3 weeks in Las Vegas playing blackjack. He’s
on the tables between 10 and 14 h each and every day, plays about 30 hands
each hour, and wagers between $2 and $10 on every hand, most frequently
toward the low end. Estimate the total amount of money he has wagered
(“action”) during his “vacation.” Note your assumptions. (Do not calculate
high and low amounts. Estimate one best value.)
28
1 A Review of Mathematical Concepts and Tools
STUDENT’S NAME ________________________________________________
33. Some believe that the oceans of the Earth were created by a long-lasting
continuous worldwide deluge. Estimate the number of years required to fill
the ocean basins of the Earth to their current depth. Note your assumptions.
34. Estimate the cost to the U.S. taxpayer when 100 United States Senators
campaign for re-election. Include taxpayer-supported services provided to
Senators that they might utilize in their re-election campaigns. Clearly note
all your assumptions and units. Compare this to the $5 million annual cost of
the Search for Extraterrestrial Intelligence (SETI) program, cancelled by the
same group. (Extra credit: Compare the likelihood of finding extraterrestrial
intelligence to the likelihood of finding intelligence in the United States
Senate).
M.
Mathematical Concepts Experiment Exercises
29
STUDENT’S NAME ________________________________________________
35. Estimate the time that Santa has available to travel to and work in each
individual home in the world that celebrates the yuletide holiday. Note your
assumptions.
Convert the following numbers into the unit requested.
36. 50 kg into grams (1 kg ¼ 1000 g)
37. 1.45 1023 m into parsecs (1 pc ¼ 3.086 1016 m)
38. 1500 m into miles (1 mi ¼ 1.61 km; 1 km ¼ 1000 m)
39. 108 spiral galaxies into solar masses (one spiral galaxy contains about 3 1011
solar-type stars)
40. A table whose length is known to be 8.40 m is measured to have a length of
8.35 m. Calculate the percentage error in the measurement.
30
1 A Review of Mathematical Concepts and Tools
STUDENT’S NAME ________________________________________________
41. Barnard’s star, a nearby star that is thought to possess a planetary system, has
an apparent visual magnitude of 9.54m. A student astronomer measures its
apparent magnitude at 9.36m. What is the percentage error in the measurement?
42. These are the average heights of a group of extraterrestrial visitors. Find their
mean and standard deviation. 17.8 m, 19.2 m, 16.3 m, 17.2 m, 16.9 m
43. Convert 27.14 into degrees, minutes, and seconds of arc.
44. Convert 41 500 into degrees and decimal fraction of a degree.
M.
Mathematical Concepts Experiment Exercises
31
STUDENT’S NAME ________________________________________________
45. Convert 7 100 5600 into degrees and decimal fraction of a degree.
46. A galaxy is known to be at a distance of 150 Mpc (1 Mpc ¼ 106 pc). It subtends
an angle of 2500 of arc. Calculate its diameter in parsecs. Compare this to the
diameter of the Milky Way galaxy, about 30,000 pc, or 30 kpc.
47. Using a ruler calibrated in millimeters and the wavelengths in angstroms
provided, determine the scale factor of the following portion of the solar
spectrum in angstroms per millimeter (Å /mm).
32
1 A Review of Mathematical Concepts and Tools
STUDENT’S NAME ________________________________________________
48. The galaxy pictured in the drawing, similar to NGC 4303, an SBc galaxy in the
Virgo cluster of galaxies, has an angular size of 110 of arc, about 1/3 the angular
diameter of the moon. (NGC is the abbreviation for “New General Catalog.”)
Determine the scale factor of the drawing in minutes of arc per millimeter (min
of arc/mm).
3 mm
NGC 4303
49. In the sketch of a star field below, we are told that the scale factor is 47 seconds
of arc per millimeter. Determine the angular distance between the stars labelled
A and B. (Most photographs in astronomy, particularly for research purposes,
are presented in the negative image, more easily analyzed than the positive
image to which we are accustomed.)
M.
Mathematical Concepts Experiment Exercises
33
STUDENT’S NAME ________________________________________________
50. The calculation to determine the velocity of impact of M31 and the Milky Way
galaxy assumed that the mass of the Milky Way was 580 billion solar masses.
Research in 2009 indicated that that Milky Way mass was larger, 710 billion
solar masses, about equal to that of M31.
a) Using (8), calculate the velocity of impact that would result if the Milky
Way has a mass of 710 billion solar masses. Show all your calculations here,
displaying all the units.
b) In supersonic flight, the term Mach number (Ernst Mach, 1838–1916) refers
to the ratio of the velocity of supersonic flight to the speed of sound in the
medium. Calculate the Mach number describing the velocity of impact of
M31 with a Milky Way of mass equal to 710 billion solar masses.
51. The Julian date corresponding to December 4, 2011, is 55899.
a. How many days will have elapsed from December 4, 2011, until June 7,
2013? Show your calculations here.
b. The Julian date corresponding to June 7, 2013, is 56450. Perform the same
calculation using Julian dates.
Removing these DATA SHEETS from the book may damage the binding. You might
consider entering the data and performing your calculations in the book, and then
photocopying the DATA SHEETS for submission to your instructor for grading.
http://www.springer.com/978-1-4614-3310-1